# If the quotient of a local ring is regular, does that imply that the original ring must be regular?

Suppose $R$ is a local ring and let $I\subset R$ be some nontrivial ideal. Are there conditions that we can place on $I$ so that if $R/I$ is regular, then so is $R$?

I am aware of the result that states: if $R$ is already a regular local ring, then $R/I$ is regular iff $I$ is generated by a subset of a regular system of parameters. But I am wondering more about whether or not the regularity of $R$ itself can be determined by it's quotient $R/I$.

Background for this question: after reading the responses to the question When is a blow-up non-singular? I am trying to work through the first argument in the second section of the paper "On the smoothness of blow-ups." I think the author uses the result that I am asking about when they state: "$S_P/f_i S_P$ is a regular local ring; since $f_i$ is a non-zero-divisor on $S_P$, it follows that $S_P$ is itself regular." I am just trying to figure out where this result comes from.

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Let $R$ be a local ring with maximal ideal $\mathfrak m$ and $x\in\mathfrak m-\mathfrak m^2$. Then $R$ is regular iff $R/(x)$ is regular and $x$ doesn't belong to any minimal prime.
@QiL'8 Maybe I'm missing something: in your case, if, for instance, $r=1$ and $R/(x)$ is regular, then the embedding dimension of $R/(x)$ equals $\dim R/(x)=\dim R-1$. I think now it follows that $x\notin\mathfrak m^2$. (What I want to say is that in the end the elements $x_i$ turn out to be outside of $\mathfrak m^2$, although you are not assuming this.) – user26857 Jul 7 '13 at 19:13